In basic algebra, collocation method is a method for transforming quadratic polynomial into the sum of the square of linear polynomial and constant. This method converts the polynomial in the following form into coefficients A, B, C, D and E in the above expression, which can also be expressions and contain variables other than X. ..

Matching method is usually used to derive the root formula of quadratic equation: our aim is to turn the left side of the equation into a complete square. Since the complete square in the problem has the form of (x+y)2 = x2+2xy+y2, 2xy = (b/a)x can be deduced, so y = b/2a.

Add y2 = (b/2a)2 to both sides of the equation, and you can get: this expression is called the root formula of quadratic equation.

Solving the equation: in the unary quadratic equation, the matching method is actually to shift the terms of the unary quadratic equation and add the square of half the absolute value of the coefficient of the first term on both sides of the equal sign.

Example solution equation: 2x? +6x+6=4

Analysis: the original equation can be arranged as: x? +3x+3=2, which can be obtained by the formula (x+ 1.5)? = 1.25 can be solved by finding the root.

Solution: 2x? +6x+6=4

& lt= & gt(x+ 1.5)？ = 1.25

X+1.5 = the square root of 65438 +0.25.

Find the maximum value

Example it is known that real numbers x and y satisfy x? +3x+y-3=0, then the maximum value of x+y is _ _ _ _.

Analysis: y is represented by a formula containing x, and then it is substituted into (x+y) for evaluation.

Solution: x? +3x+y-3 = 0 & lt； = & gty=3-3x-x？ ,

Replace x+y = 3-2x-x with (x+y)? =-(x？ +2x-3)=-[(x+ 1)？ -4]=4-(x+ 1)？ .

Because (x+ 1)? ≥0, so 4-(x+ 1)? ≤4. Therefore, it is assumed that the maximum value of (x+y) is 4. At this time, both X and Y have solutions, so the maximum value of (x+y) is 4.